evaluate lim x tends a f(x) where f(x) = (ax- xa)/( ax - xx)
Given y = (ax - xa) / (ax - xx )
As limit a → 0 , y = 0/0 which is indeterminate
Now differentiating numerator and denominator separately we get
d(xa)/dx = axa-1
for xx ,
Let k = xx
ln k = x ln x
Differentiating both sides we get
(1/k)dk/dx = (ln x )(dx/dx) + (d (lnx)/dx) (x) = ln x + (1/x)x = 1+ ln x
d(ax)/dx = ax(ln a)
y = lim ((ln a)ax - axa-1) / ((ln a)ax - (1+ln x )xx)
⇒ y = ((ln a)aa - aaa-1) / ((ln a)aa - (1+ln a )aa)
⇒ y = ((ln a)aa - aa) / ((ln a)aa - (1+ln a )aa)
= aa(ln a -1)/(aaln a -aa -aa lna)
= aa(ln a -1)/(-aa)
= 1- ln a